actuators (1)
TRANSCRIPT
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RR
ActuatorsActuators
Instructor: Dr. Hong Zhang
Mechanical Engineering
Rowan University
ME ROWANME ROWAN 22
RR Homework 1Homework 1
Chapter 2:
Problem 1, 10, 14, 17, 18
Chapter 3:
Problem 9
Due: March 2, 2004
ME ROWANME ROWAN 33
RR ActuatorsActuators
Actuators are the elements that transformthe control inputs to actions of a machine.
The common types of actuators:
Electric motors
DC & AC motors
Stepper motors
Servo motors
Hydraulic & Pneumatic actuators
Shape memory metal actuators
Others
ME ROWANME ROWAN 44
RR Mechanical Actuation SystemsMechanical Actuation Systems
Mechanisms: Devices that transform motionfrom one form to some other required form.
Mechanical elements: Linkage, cam, gear,rack-and-pinion, chain, belt drive, etc.
Functions:
Force amplification
Change of speed
Transfer rotation between axis Others
ME ROWANME ROWAN 55
RR Four Bar LinkFour Bar Link
1
4
3
2
1
4 3
2
14 3
2
Double-lever
2143 llll +>+ 2143 llll +
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ME ROWANME ROWAN 77
RR Gear TrainGear Train
A
B
B
A
n
n=
C
D
A
B
C
D
A
B
D
A
n
n
n
n
n
n
n
n=
=
A C DB
A B
ME ROWANME ROWAN 88
RR Planetary Gear TrainPlanetary Gear Train
SunRing Gear
Planet
Arm
armF
arm
gearsdrivenonteethofnumberofproduct
gearsdriveronteethofnumberofproduct
= L
ME ROWANME ROWAN 99
RR
The sun has 40 teeth, the planet 20 teeth,and the ring gear 80 teeth. The arm is inputand the sun is output. The ring gear is heldstationary. Find the train ratio of sun/arm.
ratio of sun/arm =3
ME ROWANME ROWAN 1010
RR
Build a remotely controlled mouse:
Basic parts: a mouse set & a joystick set
Other material: anything you can buy within the limit ofthe allowance (which you can keep it as secret), andanything you find from machine shop and other places(which you need to inform the entire class and getapproval).
Tool: machines and tools available in machine shop
and electric lab, you can also use your own tool Allowance: $10 (dept.) + $10 (self), keep receipts for
peer examination prior competition
Project 1: Mouse HouseProject 1: Mouse House
ME ROWANME ROWAN 1111
RR
Date: March 10, 2004
Mouse soccer tournament Time: 5 minutes each game for 3 games
You can design and install any ball handling tool
Sumouse triangle: Time: 2 minutes
Push or trick the others out of the boundary.
Point = 10 # of stepping out down time (every 5s)
You can add any attachment to enhance your chance
CompetitionCompetition
ME ROWANME ROWAN 1212
RR Characteristics of ActuatorsCharacteristics of Actuators
Weight
Power-to-weight ratio
Stiffness: The resistance of a materialagainst deformation.
Stiff vs.Compliant
Accuracy vs. Flexibility
Reduction gears
Reduce speed, increase torque
Increased cost, number of parts, backlash, etc.
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ME ROWANME ROWAN 1313
RR Bucket ScienceBucket Science
You have a 5L and a 3L bucket and unlimitedsupply of water. How to use these twobuckets to measure any amount of water
from 1L to 8L.
5L 3L
ME ROWANME ROWAN 1414
RR Hydraulic ActuatorsHydraulic Actuators
High power- to- weight ratio (depending on whetherthe pump & reservoir are included)
Large forces at low speeds (linear or rotary)
Compatible with microprocessor and electronic
control No reduction gears required
Tolerance of extreme hazardous environments
May leak
Expensive and noisy, requires maintenance
Depends on the properties of oil (dirt, temp., etc)
ME ROWANME ROWAN 1515
RR ApplicationsApplications
Dump Truck
Car Brakes
Shovel
ME ROWANME ROWAN 1616
RR Principle of Hydraulic SystemsPrinciple of Hydraulic Systems
F F
Oil
Pistons have equal areasF F
Oil
Pistons with different areas
F=F
21
'
A
F
A
F=
FA
AF
1
2'=
A1 A2
ME ROWANME ROWAN 1717
RR Common ComponentsCommon Components
Power source
Hydraulic linear or rotary cylinders and rams
Hydraulic switches and other controllers
Connecting hoses
Safety valves and other mechanisms
Sensors
ME ROWANME ROWAN 1818
RR Schematic of a Hydraulic SystemSchematic of a Hydraulic System
Sensors Controller
HydraulicPowerSource
Switch
Return
Source
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ME ROWANME ROWAN 1919
RR Power SourcePower Source
Reservoir
Hydraulic pump operated by a motor or an
engine Cooling system (for pneumatic system)
ME ROWANME ROWAN 2020
RR Power SourcePower Source
Return
Oil Reservoir
Pump
Motor
None-return
valve
Pressure
relieve valveAccumulator
To system
ME ROWANME ROWAN 2121
RR Hydraulic CylindersHydraulic Cylinders
Hydraulic cylinder
Telescope cylinderSchematic drawing
ME ROWANME ROWAN 2222
RR Directional Control ValvesDirectional Control Valves
Port 1Port 3 Port 2
Spool
Port 3 Port 1Port 2
One-way valve
ME ROWANME ROWAN 2323
RR Valve SymbolsValve Symbols
Flow path Flow shut-off InitialConditions
Push-buttonSpring-button
RollerSolenoid Pneumatic Lever Pedal
ME ROWANME ROWAN 2424
RR Example: Poppet ValveExample: Poppet Valve
Poppet valve
2 port 2 position (2/2) poppet valve
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ME ROWANME ROWAN 2525
RR More ExamplesMore Examples
3/2 valve 4/2 valve
Lift system
Vent
Pressure supply
ME ROWANME ROWAN 2626
RR 22--Actuator Sequential OperationActuator Sequential Operation
5
6
7
b+b-Cylinder BLimit switches
a-
a+
Start1
4
3
2
a+a-Cylinder ALimit switches
b-b+
ME ROWANME ROWAN 2727
RR Brain SugarBrain Sugar
Use 1, 5, 6, 7 and any combination of +, -, *and / to obtain 24 and 21.
2465*)17( =
21)7/51/(6 =
ME ROWANME ROWAN 2828
RR Electric MotorsElectric Motors
Most popular motor
Types:
DC motor
AC motor
Brushless motor
Step motor
Servo motor
Linear motor
Applications:
ME ROWANME ROWAN 2929
RR Basic Principles of Electromagnetic MotorsBasic Principles of Electromagnetic Motors
N
S
B
1. A magnetic field B is generated by a pair of magnets.
2. A conductor with no current will not disturb the field B.
x
3. When a conductor carries an electrical current, aconcentric magnetic field is produced along theconductors length.
4. Combined action between the two magnetic fieldscreate a mutual repulsive force forcing thewire downwards.
F
x B
BilFrr
=
ME ROWANME ROWAN 3030
RR
When there is a loop of conductor carrying current inthe magnetic field, a torque will be generated.
Electromagnetic MotorElectromagnetic Motor
Current
N S
B
F
F
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ME ROWANME ROWAN 3131
RR Brushes and CommutatorBrushes and Commutator
For continuous operation, a pair of brushes andcommutator are used to change the direction of current
ME ROWANME ROWAN 3232
RR Direction Changing w/ BrushesDirection Changing w/ Brushes
(a) (b)Armature
ME ROWANME ROWAN 3333
RR A Simple MotorA Simple Motor
ME ROWANME ROWAN 3434
RR Inside a DC MotorInside a DC Motor
Field magnet
DC motor
Armature
Commutator
ContactBrushes
ME ROWANME ROWAN 3535
RR DiscussionDiscussion
How does this worlds simplest motor work?
How to reverse the direction of rotation?
ME ROWANME ROWAN 3636
RR Field CoilsField Coils
Practical motors have several loops on anarmature to provide a more uniform torqueand the magnetic field is produced by anelectromagnet arrangement called the fieldcoils.
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ME ROWANME ROWAN 3737
RR Arrangement of Field CoilsArrangement of Field Coils
Field coil
Armature coil
Series wound motor
Field coil
Armaturecoil
Shunt wound motor
Field coil
Armaturecoil
Field coil
Compound motor
Armaturecoil
Field coil
Separately wound motor
ME ROWANME ROWAN 3838
RR Torque with Permanent MagnetTorque with Permanent Magnet
Force generated by oneconductor
BilF=
i
B
F
F
l
b
Torque generatedbBilT =
iknbBliT t ==
Total torque
ME ROWANME ROWAN 3939
RR Torque with Field CoilTorque with Field Coil
For motor with field coil, we just need tochangeB toKfIstator.
ikliIbKnT tstatorf ==
ME ROWANME ROWAN 4040
RR Back ElectroBack Electro--Magnetic FieldMagnetic Field
Rotating armature coil in magnetic field willinduce back e.m.f. vb
vb kv =
V
R L
Vb
A DC motor can be considered as anequivalent circuit
RkV
RvVi vb
=
=i
ME ROWANME ROWAN 4141
RR Torque and SpeedTorque and Speed
Ignoring the inductance of armature, the torque is
Where R is the armature resistance, V is theapplied voltage, and the rotational speed.
R
kVkikT vtt
)( ==
Torque
Rotational speed
V1
V2
V3
ME ROWANME ROWAN 4242
RR Torque Characteristics of a MotorTorque Characteristics of a Motor
Rotation rate (rpm)
Torque
(Nm)
10 60504030200
0.1
0.5
0.4
0.3
0.2
0.8
0.7
0.6
A 12V DC motor
A 24V DC motor
Stall
No load
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ME ROWANME ROWAN 4343
RR Brain SugarBrain Sugar
Add following lines together
4100
1000
20
1030
1020
20
+) 1010
ME ROWANME ROWAN 4444
RR
Connect all 9 dots with 4 connected straight lines
ME ROWANME ROWAN 4545
RR Design ProblemDesign Problem
Known: Total mass m= 5 kg. We have 10 AA NiCd batterieswith 1.2V, 500mAh, and a DC motor with normal runningcurrent 0.15A at 12V. Suppose transmission ratio 1:10 and
stall torque of the motor is 300 oz.in.Check: 1. Whether the car can reach acceleration a=0.1g?
2. How long will the battery last?
Body
Wheel (r=4cm)
MotorBattery
Transmission
ME ROWANME ROWAN 4646
RR CalculationCalculation
Acceleration: a = 0.1 x 9.8 =0.98 m/s2
Force: F = ma = 5 x 0.981 = 4.9 N
Wheel torque:
Tw= Fr = 4.9 x 0.04 = 0.196 Nm
Minimum torque required from the motor:
Tm= Tw*10 = 1.96 Nm
= 141.61*1.96 oz.in = 278 oz.in
Stall torque of the motor is 300 oz.in300 oz.in > 278 oz.in
ME ROWANME ROWAN 4747
RR CalculationCalculation
Running power of the motor:P=VI= 12 x0.15 = 1.8 w
Running time: 500mAh = 0.5 Ah
0.5 / 0.15 = 3.3 hr
Note: You need to consider other factors inreal design, such as friction in transmission,uneven surface, etc.
ME ROWANME ROWAN 4848
RR DiodeDiode
Diode
Cathode
AnodeV
I
Forwardbiased
Reversebiased
0
Diode characteristic
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ME ROWANME ROWAN 4949
RR Bipolar TransistorBipolar Transistor
BFEcIhI =
Ic Ic
Base current
Ic
Base current
Ic
Emitter
Base
Collector
NPN
Collector
Emitter
Base
PNP
ME ROWANME ROWAN 5050
RR Darlington PairsDarlington Pairs
NPN PNP
IcIc
Base current
Ic
Base current
IcIc Ic
ME ROWANME ROWAN 5151
RR H Circuit DC Motor ControllerH Circuit DC Motor Controller
V+
High
Low
ME ROWANME ROWAN 5252
RR H Circuit DC Motor ControllerH Circuit DC Motor Controller
V+
High
Low
ME ROWANME ROWAN 5353
RR MOSFETMOSFET
Drain
Source
Gate
Drain
Source
Gate
12V
12V
GND
Output portof micro-
processor
N-channel P-channel
1.High frequencyswitching
2.Easy interfacing tomicroprocessor
ME ROWANME ROWAN 5454
RR Electronic Speed ControllerElectronic Speed Controller
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ME ROWANME ROWAN 5555
RR Commercial ESCCommercial ESC
ME ROWANME ROWAN 5656
RR Mirror TimeMirror Time
There are times when hour and minutereadings are same, such as 1:01, 2:02, 3:03,etc. What is the shortest possible timebetween two different readings of this mirrortime?
49min, 12:12 1:01
ME ROWANME ROWAN 5757
RR Who is lying?Who is lying?
Four friends Andrew, Barbara, Cindy, andDaniel were shown a number. Here is whatthey had to say about that number:
Andrew: It has two digits
Barbara: It goes evenly into 150
Cindy: It is not 150
Daniel: It is divisible by 25
It turns out that one and only one of the fourfriends is lying. Which one is it?
D
ME ROWANME ROWAN 5858
RR Stepper MotorsStepper Motors
A motor produces rotation through equalangles, the so-called steps, for each digitalpulse supplied to its input.
NS
N S
N
S
N
S
ME ROWANME ROWAN 5959
RR Obtain Smaller StepsObtain Smaller Steps
A stepper motor with 4 teeth rotor and 6 teeth stator willobtain minimum step of 15
(A)
1
23
4
5 6
a
b
c
d
(C)
1
23
4
5 6
ab
cd
(B)
1
23
4
5 6
a
b
c
d
Develop a control scheme for a stepper motor with 8teeth rotor and 6 teeth stator
Given a 50 teeth rotor and 40 teeth stator, what isthe minimum step?
ME ROWANME ROWAN 6060
RR MicroMicro--SteppingStepping
In micro- stepping, the
power- on or power- off for each coil is donegradually by dividingthe changes intosmaller divisions.
N
S
N
S
N
S
Half stepping
Pro and Cons.
Smaller stepping angles
Smoother rotation
Complex controller
Develop half stepping scheme for the previous 4/6 example
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ME ROWANME ROWAN 6161
RR Servo MotorServo Motor
A servomotor is any motor with feedback that canbe controlled to move at a desired speed (andconsequently, torque), for a desired angle ofrotation.
MotorDynamics Load
Differentiate
Feedback
Desiredposition &velocity +
_Position
Velocity
Error
ME ROWANME ROWAN 6262
RR Anatomy of a Servo MotorAnatomy of a Servo Motor
ME ROWANME ROWAN 6363
RR Angle ControlAngle Control
ME ROWANME ROWAN 6464
RR Digital Control of a MotorDigital Control of a Motor
The speed of the motor is controlled byvarying voltage
Control signal from computer is digital, henceneed conduct digital-to-analog converters(DAC)
The resolution of the control is determined bythe number of ports or bits (2n)
ME ROWANME ROWAN 6565
RR Servo Motor ControllerServo Motor Controller
PIC16C73 based
SV203 Servo
Motor Controller
PRINT #1,"BD";ID$;"SV";Servo$;"M";Position$
ME ROWANME ROWAN 6666
RR Pulse Width Modulation (PWM)Pulse Width Modulation (PWM)
The voltage is sent as pulse with constant voltage
The average voltage is determined by the width ofthe pulse
Time
V
tT
Vcc
Vout
T
tVV ccout
=
T 4T3T2T 5T0
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ME ROWANME ROWAN 6767
RR Industrial AC MotorIndustrial AC Motor
1000 HP1 HP
ME ROWANME ROWAN 6868
RR AC Induction MotorAC Induction Motor
Same theory of DC motor to turn the rotor
No wiring required for rotor
3-phase AC to induct rotating electro-
magnetic field.
ME ROWANME ROWAN 6969
RR Component of an AC MotorComponent of an AC Motor
Stator Rotor
ME ROWANME ROWAN 7070
RR 3 Phase Alternating Current (AC)3 Phase Alternating Current (AC)
ME ROWANME ROWAN 7171
RR 3 Phase AC Wiring3 Phase AC Wiring
iA
iC
iBD2
D1
D3
D5D4
D6
Y (Star) wiring
(Triangle) wiring
ME ROWANME ROWAN 7272
RR
D3
D6
D1
D4
D5
D2
iA D1
iC
iBD2
D3
D5 D4
D6
iA
iB
iC
V
t
0
t1 t2 t3 t4 t5 t6
t= 0x
x
N
S
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ME ROWANME ROWAN 7373
RR Turning ElectroTurning Electro--Magnetic FieldMagnetic Field
D3
D6
D1
D4
D5
D2
iA D1
iC
iBD2
D3
D5 D4
D6
iA iBiCV
t
0
t1 t2 t3 t4 t5 t6
t= t1, t2,?
???
ME ROWANME ROWAN 7474
RR
D3
D6
D1
D4
D5
D2
x
x
x
D3
D6
D1
D4
D5
D2
x
x
t= t1t= t2
t= t3
t= t4
D3
D6
D1
D4
D5
D2x
x
x
D3
D6
D1
D4
D5
D2
x
x
x
D3
D6
D1
D4
D5
D2x
x
x
t= t5 t= t6
D3
D6
D1
D4
D5
D2x
x
x
ME ROWANME ROWAN 7575
RR Use of Reduction GearsUse of Reduction Gears
In most motor powered system, we use reductiongears or other mechanics to reduce speed andincrease torque.
Inertia of loadIl, Inertia of motorIm, Gear ratio N,viscous friction of load and motor are IlandIm
Motor
Gears
Loadm& L
&
Tm
ME ROWANME ROWAN 7676
RR Dynamic EquationsDynamic Equations
mNl && 1= mNl
&&&& 1=and
lN
m
l TT1=
mlNmmlNm
llllNmmmm
lNmmmmm
bbII
bIbI
TbIT
&&&
&&&&&&
&&&
)()(
)(
22
11
1
1
+++=
+++=
++=
Total inertia Total friction
ME ROWANME ROWAN 7777
RR ExerciseExercise
A motor withIm=0.015 kgm2 and maximum torque of 8Nm is
connected to a uniformly distributed arm with aconcentrated mass at its end. Ignoring the inertia felt by themotor and viscous friction in the system, calculate the totalinertia felt by the motor and the maximum angularacceleration it can develop if the gear ratio is (a) N=3 and(b) N=30.
Motor
m& L
&
0.5m
3kg
2kg